huangyhg

gr. 8.8 bolt strength
can someone tell mee the strength of gr 8.8 bolts (bs 3692)? i'm looking for ultimate, yield, and shear strength.
thanks,
chris

hi meca
the grade 8.8 tells you the yield strength and ultimate stress, it works like this:-
multiply 8*8 then * 10=640n/mm^2 yield stress
multiply 8.8 *100 = 880n/mm^2 tensile stress
regards desertfox
hi meca
sorry the last equation should have been the first number 8
multiplied by 100 to give the ultimate tensile strength
ie:- 8*100= 800n/mm^2
both these values are ie yield and ultimate tensile are minimum values.
in addition the yield stress is the stress at a permanent
set limit of 0.2%
regards desertfox
thanks desertfox. do you know what is typically used for the allowable shear stress (single plane)?
hi meca
it is normal to assume about 70% of the ultimate tensile stress for the shear stress.
regards desertfox
desert fox
is the mm^2 the nominal bolt dia, or root dia of the thread
regards
pat
eng-tips, by professional engineers for professional engineers
hi patprimmer
i haven't got the correct bs spec in front of me now, but
yes i think its based on the root diameter of the thread as
that would be conservative for design calculations.
in fact all the design calculations relating to tension that i have seen are based on minimum root diameter.
regards desertfox
an 8.8 bolt has an ultimate tensile strength of 80 kg/mm2 (785n/mm2) and a yield strength of 80% x 80kg/mm2 = 64 kg/mm2 (628n/mm2)
surely an allowable shear stress of 70% x ultimate tensile stress is much too high. this 550 n/mm2.
generally, the shear stress is equal to 60% x yield stress. this gives a shear stress of 375 n/mm2. this stress only applies to factored loads.
if using unfactored loads, i would use half this valus, 185 n/mm2.
regards
hi meca
linslade is correct the value i quoted would be the ultimate shear stress not the normal working shear stress.
regards desertfox
in general, twice the shear stress is equivalent to the stress intensity, though that definition might not be quite correct. the stress intensity is limited by the yield stress and hence most design codes limit the shear stress to half the yield stress. i'd check with a design standard on correct factors to apply.
corus

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